A 1.00 cm-high object is placed 3.40 cm to the left of aconverging lens of focal length 8.95 cm. A diverging lens of focallength −16.00 cm is 6.00 cm to the right of the converging lens.Find the position and height of the final image.
position
Take the image formed by the first lens to be the object for thesecond lens and apply the lens equation to each lens to locate thefinal image. cm  ---Select--- behind the second lens infront of the second lens height
Calculate the magnification produced by each lens. Then considerhow the magnification relates image size and object size for eachlens to find the height of the final image. cm
Is the image inverted or upright?
uprightinverted   Â
Is the image real or virtual?
realvirtual   Â
(b)
What If? If an image of oppositecharacteristic, i.e., virtual if the image in part (a) is real andreal if the image in part (a) is virtual, is to be obtained, whatis the minimum distance (in cm), and in which direction, that theobject must be moved from its original position?
distance cmdirection ---Select--- to the left to the right
